Final Answer:
The pH of the resulting solution is approximately 11
Step-by-step explanation:
To analyze the pH of the resulting solution, we need to consider the reactions and the resulting equilibrium:
1. Dissolution:
H₂CO₃ (carbonic acid) partially dissolves in water to form HCO₃- (bicarbonate) and H+ (hydrogen ions):
H2CO₃ + H₂O ⇌ HCO₃- + H+
NaHCO₃ (sodium bicarbonate) dissolves completely in water to form Na+ and HCO₃- ions:
NaHCO3 ⇌ Na+ + HCO3-
Na₂CO₃ (sodium carbonate) dissolves completely in water to form 2Na+ and CO₃⁻²- (carbonate) ions:
Na₂CO₃ ⇌ 2Na+ + CO₃²-
NaOH (sodium hydroxide) dissociates completely in water to form Na+ and OH- (hydroxide) ions:
NaOH ⇌ Na+ + OH-
2. Equilibrium:
Water itself undergoes autoionization, forming H+ and OH- ions:
H2O ⇌ H+ + OH-
3. Acid-base reactions:
H₂CO₃ and HCO₃- participate in a buffer equilibrium:
H₂CO₃ + H₂O ⇌ HCO₃- + H+
OH- ions from NaOH react with H+ ions to produce water:
H+ + OH- ⇌ H₂O
4. Determining pH:
To calculate the pH, we need to understand the dominant species present in the solution. We can analyze the initial concentrations:
- Initial [H₂CO₃] = 10² M
- Initial [NaHCO₃] = 10⁻² M
- Initial [Na₂CO₃] = 10⁻² M
- Initial [NaOH] = 10⁻² M
Considering the complete dissociation of NaHCO₃, Na₂CO₃, and NaOH, the final concentrations of Na+, HCO₃-, CO₃²-, and OH- will be much higher than the initial concentration of H₂CO₃. Therefore, H₂CO₃ will contribute very little to the [H+] concentration.
Since the solution contains both OH- ions from NaOH and a buffer system from H₂CO₃/HCO₃-, we can expect a basic solution. The pOH will be determined by the remaining OH- ions after reacting with some H+ ions from H₂CO₃.
Calculating pOH:
1. Calculate the initial moles of OH- from NaOH:
moles OH- = 0.1 L * 0.01 mol/L = 10⁻³ mol
2. Calculate the moles of H+ neutralized by OH- using the reaction stoichiometry:
moles H+ = moles OH- = 10⁻³ mol
3. Now, consider the buffer equilibrium of H₂CO₃/HCO₃-:
Ka = [H+][HCO₃-]/[H₂CO₃]
Where Ka is the acid dissociation constant of H₂CO₃ (4.3 x 10⁻⁷).
4. Substitute the initial concentration of H₂CO₃ and use the fact that [H+] and [HCO₃-] are equal at equilibrium:
Ka = [H+]² / [H₂CO₃] = [H+]² / 10⁻² M
[H+] = sqrt(Ka * [H₂CO₃]) = sqrt(4.3 x 10⁻⁷ * 10⁻² M) = 2.07 x 10⁻⁵ M
5. Calculate the remaining moles of H+ after neutralization:
moles H+ remaining = 10⁻² mol - 10⁻³ mol = 9 x 10⁻⁴ mol
6. Calculate the final concentration of H+ and pOH:
[H+] = moles H+ remaining / volume = 9 x 10⁻⁴ mol / 0.1 L = 9 x 10⁻⁵ M
pOH = -log[OH-] = -log[10⁻³ mol / 0.1 L] = 3
Calculating pH:
pH = 14 - pOH = 14 - 3 = 11
Therefore, the pH of the resulting solution is approximately 11.